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JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Afternoon)

  • question_answer
    5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If \[{{C}_{V}}=28\,J{{K}^{-1}}mo{{l}^{-1}},\]calculate \[\Delta U\]and \[\Delta pV\] for this process. \[(R=8.0J{{K}^{-1}}mo{{l}^{-1}})\]             [JEE Main 8-4-2019 Afternoon]

    A) \[\Delta U=14\,kJ;\,\,\,\,\,\Delta (pV)=4\,kJ\]

    B) \[\Delta U=14\,kJ;\,\,\,\,\,\Delta (pV)=18\,kJ\]

    C) \[\Delta U=2.8\,kJ;\,\,\,\,\,\Delta (pV)=0.8\,kJ\]

    D) \[\Delta U=14\,kJ;\,\,\,\,\,\Delta (pV)=0.8\,kJ\]

    Correct Answer: A

    Solution :

    \[n=5;\,\,\,\,\,{{T}_{i}}=100K;\,\,\,\,\,{{T}_{f}}=200K;\]           \[{{C}_{V}}=28\,J/mol\,K;\,\,\,\,\,\,\,\,\,\,Ideal\,gas\]           \[\Delta U=n{{C}_{V}}\Delta T\] \[=5\,mol\,\times 28\,J/mol\,K\times (200-100)K\] \[=14,000J=14kJ\] \[\Rightarrow \]\[{{C}_{p}}={{C}_{v}}+R=(28+8)J/mol\,K\] \[=36\,J/mol\,K\] \[\Rightarrow \]\[\Delta H=n{{C}_{p}}\Delta T=5\,mol\times 36J/mol\,K\times 100K\] \[=18000J\,=18\,kJ\] \[\Delta H=\Delta U+\Delta (PV)\] \[\Rightarrow \]\[\Delta (PV)=\Delta H-\Delta U=(18-14)kJ\,=4\,kJ\]


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